Hambleton et al. (1991) suggest using a chi-square test to identify local independence between two items. The procedure consists of constructing a 2x2 table for two items using the correct and incorrect answers of persons at the same level of ability (i.e., with the same raw scores on the test):
|YR = A+B|
|Total||XR = A+C||N-XR = B+D||N = A+B+C+D|
This yields a chi-square statistic with 1 degree of freedom (without Yates' continuity correction, but the principle is the same):
This is minimized to zero, and local independence is apparently assured in this dataset at this ability level for this pair of items, when
AD = BC
A/N = XR/N * YR/N
as it would be when the data exactly conform to the Rasch model, or any other model in which the two items are conditionally independent.
In conventional usage, however, when χ2 < 3.84, there is a presumption of local independence with 95% confidence. Then "local independence" is declared!
Let us conduct an experiment based on this convention. Let us constrain the possible outcomes so that, if a subject succeeds on item X, that subject cannot fail on item Y. So the items are always locally dependent. The data matrix becomes:
|YR = A+B|
|Total||XR = A||N-XR = B+D||N = A+B+D|
The chi-square test apparently becomes:
Then we can determine the relationship between successes on items X and Y for any value of chi-square, say N*k, where k is a constant. Accordingly,
Rewriting this in terms of probabilities, where PX is the probability of success on item X under these conditions, etc., and taking logarithms, such that K=loge(k):
It is seen that the relationship between the items is expressed in terms of log-odds difficulties, in accord with the Rasch model, despite that fact that the data do not accord with Rasch model conditions (because the items are locally dependent). Item Y is always easier than item X.
Further, for any particular chi-square "significance" value, e.g., 3.84, items with paired log-odds difficulties differing by less than loge (3.84 / N) might be unwittingly declared "locally independent".
Here is an example:
The log-odds difficulties of the two items are loge(13/87) = -1.90, and loge(80/20) = 1.39 logits, differing by -3.29. The criterion value is loge(3.84/100) = -3.26. Since -3.29 is less than -3.26, the items might be misleadingly declared "locally independent". For comparison, the chi-square computation above gives: χ² = 100 * 260^2 / (87 * 13* 20 * 80) = 3.74 with 1 d.f. one-sided → p=.053, so that the null hypothesis of local independence is not rejected.
Hambleton R.K., Swaminathan H., Rogers H.J. (1991) "Fundamentals of item response theory", Sage publications Inc. London, Chapter 2, pp 9-12 and Exercise 6, pp 29-31
Chi-square local independence meets the Rasch model. Tristan A. 16:1 p.861
Chi-square local independence meets the Rasch model. Tristan A. Rasch Measurement Transactions, 2002, 16:1 p.861
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